See here. What does it mean for a prime ideal to split completely?
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If $L/K$ is a field extension and $\mathfrak{p}$ a prime of $K$, then in $L$ we have $$\mathfrak{p}\mathcal{O}_L = \prod_{i=1}^g \mathfrak{P}_i^{e_i}$$ where the $\mathfrak{P}_i$ are primes of $\mathcal{O}_L$ and $e_i\ge 1$ are integers. $\mathfrak{p}$ is totally split if all of the $e_i=1$ and if, further, for each $\mathfrak{P}_i$, the residue field degree $$f_i = [\mathcal{O}_L/\mathfrak{P}_i : \mathcal{O}_K/\mathfrak{p}] = 1.$$ Note that this means that $g = [L:K]$.
rogerl
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Don't you mean $f_i = [\mathcal{O}_?/\mathfrak{B}_i : \mathcal{O}_K/\mathfrak{p}] = 1$...? – user166854 May 29 '15 at 02:10
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@user141494 thanks. – rogerl May 29 '15 at 13:49